Problem: You have found the following ages (in years) of all 5 turtles at your local zoo: $ 51,\enspace 30,\enspace 37,\enspace 88,\enspace 63$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{51 + 30 + 37 + 88 + 63}{{5}} = {53.8\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $51$ years $-2.8$ years $7.84$ years $^2$ $30$ years $-23.8$ years $566.44$ years $^2$ $37$ years $-16.8$ years $282.24$ years $^2$ $88$ years $34.2$ years $1169.64$ years $^2$ $63$ years $9.2$ years $84.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{7.84} + {566.44} + {282.24} + {1169.64} + {84.64}} {{5}} $ $ {\sigma^2} = \dfrac{{2110.8}}{{5}} = {422.16\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{422.16\text{ years}^2}} = {20.5\text{ years}} $ The average turtle at the zoo is 53.8 years old. There is a standard deviation of 20.5 years.